Likelihood Ratio Test and Wald test provide different conclusion for glm in R

Question

I'm reproducing an example from Generalized, Linear, and Mixed Models. My MWE is below:

Dilution <- c(1/128, 1/64, 1/32, 1/16, 1/8, 1/4, 1/2, 1, 2, 4)
NoofPlates <- rep(x=5, times=10)
NoPositive <- c(0, 0, 2, 2, 3, 4, 5, 5, 5, 5)
Data <- data.frame(Dilution,  NoofPlates, NoPositive)

fm1 <- glm(formula=NoPositive/NoofPlates~log(Dilution), family=binomial("logit"), data=Data)
summary(object=fm1)

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Output

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Call:
glm(formula = NoPositive/NoofPlates ~ log(Dilution), family = binomial("logit"), 
    data = Data)

Deviance Residuals: 
     Min        1Q    Median        3Q       Max  
-0.38326  -0.20019   0.00871   0.15607   0.48505  

Coefficients:
              Estimate Std. Error z value Pr(>|z|)
(Intercept)      4.174      2.800   1.491    0.136
log(Dilution)    1.623      1.022   1.587    0.112

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 8.24241  on 9  degrees of freedom
Residual deviance: 0.64658  on 8  degrees of freedom
AIC: 6.8563

Number of Fisher Scoring iterations: 6

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Code

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anova(object=fm1, test="Chisq")

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Output

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Analysis of Deviance Table

Model: binomial, link: logit

Response: NoPositive/NoofPlates

Terms added sequentially (first to last)


              Df Deviance Resid. Df Resid. Dev Pr(>Chi)   
NULL                              9     8.2424            
log(Dilution)  1   7.5958         8     0.6466  0.00585 **
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Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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Code

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library(aod)
wald.test(b=coef(object=fm1), Sigma=vcov(object=fm1), Terms=2)

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Output

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Wald test:
----------

Chi-squared test:
X2 = 2.5, df = 1, P(> X2) = 0.11

Estimated coefficients are perfectly matching with the results given in the book but SE's are far apart. Based on LRT test the slope is significant but based on Wald and Z-test slope coefficient is insignificant. I wonder if I miss something basic. Thanks in advance for your help.

Answer 1

The main problem is that if you're going to use the ratio as your response variable, you should be using the weights argument. You must have ignored a warning about "non-integer #successes in a binomial glm" ...

Dilution <- c(1/128, 1/64, 1/32, 1/16, 1/8, 1/4, 1/2, 1, 2, 4)
NoofPlates <- rep(x=5, times=10)
NoPositive <- c(0, 0, 2, 2, 3, 4, 5, 5, 5, 5)
Data <- data.frame(Dilution,  NoofPlates, NoPositive)


fm1 <- glm(formula=NoPositive/NoofPlates~log(Dilution),
     family=binomial("logit"), data=Data, weights=NoofPlates)

coef(summary(fm1))
##               Estimate Std. Error  z value     Pr(>|z|)
## (Intercept)   4.173698  1.2522190 3.333042 0.0008590205
## log(Dilution) 1.622552  0.4571016 3.549653 0.0003857398

anova(fm1,test="Chisq")
##               Df Deviance Resid. Df Resid. Dev  Pr(>Chi)    
## NULL                              9     41.212              
## log(Dilution)  1   37.979         8      3.233 7.151e-10 ***

The LRT and Wald test results are still quite different ($p$-values of $4 \times 10^{-4}$ vs. $7 \times 10^{-10}$), but for practical purposes we can go ahead say they're both strongly significant ... (In this case (with a single parameter), aod::wald.test() gives exactly the same p-value as summary().)

The Wald vs profile confidence intervals are also moderately different, but whether CIs [shown below] of (0.7,2.5) (Wald) and (0.9, 2.75) (LRT) are practically different depends on the particular situation.

Wald:

confint.default(fm1)
##                   2.5 %   97.5 %
## (Intercept)   1.7193940 6.628002
## log(Dilution) 0.7266493 2.518455

Profile:

confint(fm1)
##                   2.5 %   97.5 %
## (Intercept)   2.2009398 7.267565
## log(Dilution) 0.9014053 2.757092